Let f(x) = begin{cases} frac{ax^{2}+2ax+3}{4x | vedclass

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(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$, the limit $\lim_{x 	o -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.Since the

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(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$, the limit $\lim_{x 	o -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.Since the denominator $(2x-1)(2x+3)$ approaches $0$ as $x 	o -\frac{3}{2}$, the numerator $ax^2 + 2ax + 3$ must also approach $0$.$a(-\frac{3}{2})^2 + 2a(-\frac{3}{2}) + 3 = 0$ $\Rightarrow \frac{9a}{4} - 3a + 3 = 0$ $\Rightarrow -\frac{3a}{4} = -3$ $\Rightarrow a = 4$.Substituting $a=4$, $f(x) = \frac{4x^2+8x+3}{(2x-1)(2x+3)} = \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \frac{2x+1}{2x-1}$ for $x 
eq -\frac{3}{2}, \frac{1}{2}$.Now, $f(f(x)) = f\left(\frac{2x+1}{2x-1}ight) = \frac{2(\frac{2x+1}{2x-1}) + 1}{2(\frac{2x+1}{2x-1}) - 1} = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3}$.Given $f(f(x)) = \frac{7}{5}$, we have $\frac{6x+1}{2x+3} = \frac{7}{5}$.$5(6x+1) = 7(2x+3)$ $\Rightarrow 30x + 5 = 14x + 21$ $\Rightarrow 16x = 16$ $\Rightarrow x = 1$.

Let $f(x) = \begin{cases} \frac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to:

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